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The area of the closed region bounded by the equation $\mid x \mid+\mid y \mid= 2$ in the two-dimensional plane is

1. $4\pi$
2. $4$
3. $8$
4. $2\pi$

Given that, $|x|+ |y| = 2 \quad \longrightarrow (1)$

Now, we can draw the figure.

The $ABCD$ is a rhombus with  $AC$ and $BD$ diagonals whose length are $4.$

Area of rhombus $ABCD = \frac{ AC\times BD}{2} = \frac{4 \times 4}{2} = 8\; \text{units}^{2}.$

Correct Answer $: \text{C}$

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